Hi,
With this question, when I'm working it out, what am I meant to do to know whether the square root of y is + or - in the graph?
Thanks!
If you had to go from \(x = y^2\) to \(y = \pm \sqrt{x}\), you'd identify that the top branch is \(y=\sqrt{x}\) and the bottom branch is \(y=-\sqrt{x}\). This is because above the \(x\)-axis, the \(y\)-coordinates are positive, and below the \(x\)-axis, the \(y\)-coordinates are negative.
Similarly, if you had to go from \(y=x^2\) to \(x=\pm \sqrt{y}\), you'd identify that the left branch is \(x=\sqrt{y}\) and the right branch is \(x=-\sqrt{y}\). This is because to the right of the \(y\)-axis, the \(x\)-coordinates are positive, and to the left of the \(x\)-axis, the \(y\)-coordinates are negative.
Same goes here. You would take \(\boxed{x-2 = -\sqrt{y}}\) and hence \(x = 2-\sqrt{y} \), because the expression \(x-2\) will be negative for values to the left of the line \(x=2\).
hey, having a bit of a 'moment'...
Find the arc length, correct to 2 decimal places, given radius is 5.9cm and angle subtended is 23degrees 12minutes...
I understand the process of l=r.theta , but when I punch in the degrees bit into my calculator, the answer is way off!
That formula only works when \(\theta\) is radians. Write the angle as \( \left( 23 + \frac{12}{60} \right) \) degrees, and hence convert it to radians by writing it as \( \left( 23 + \frac{12}{60} \right) \times \frac\pi{180} \).